Program Explanation Given information: The system is x ′ = [ 1 0 0 − 2 − 2 − 3 2 3 4 ] x . Explanation: Consider matrix form of the system of equation as A . A = [ 1 0 0 − 2 − 2 − 3 2 3 4 ] The characteristics equation of the coefficient matrix is, [ 1 − λ 0 0 − 2 − 2 − λ − 3 2 3 4 − λ ] = 0 ( 1 − λ ) [ − 2 − λ − 3 3 4 − λ ] + 0 + 0 = 0 − λ 3 + 3 λ 2 − 3 λ+1 = 0 Further solve the above equation as shown below. − λ 3 + 3 λ 2 − 3 λ + 1 = 0 ( λ- 1 ) 3 = 0 λ = 1 , 1 , and 1 The Eigen vector of the coefficient matrix A is in the form as, [ 1 − λ 0 0 − 2 − 2 − λ − 3 2 3 4 − λ ] [ a b c ] = [ 0 0 0 ] The associated Eigen vector with the matrix A is v = [ a b c ] T . Substitute 1 for λ in coefficient matrix, [ 1 − 1 0 0 − 2 − 2 − 1 − 3 2 3 4 − 1 ] [ a 1 b 1 c 1 ] = [ 0 0 0 ] [ 0 0 0 − 2 − 3 − 3 2 3 3 ] [ a 1 b 1 c 1 ] = [ 0 0 0 ] The scalar equation are shown below, − 2 a 1 − 3 b 1 − 3 c 1 = 0 2 a 1 + 3 b 1 + 3 c 1 = 0 Consider k 1 as c 1 , k 2 as b 1 then the value of a 1 can be obtained as, a 1 = − 3 2 k 2 − 3 2 k 1 Therefore the associated Eigen vector is obtained as: u 1 = [ 3 0 − 2 ] Similarly the second case the associated Eigen vector is shown below: u 2 = [ 3 − 2 0 ] The one Eigen vector is missing, therefore the defect is 1. The other two associated vectors can be obtained by the given formula, v 1 = ( A − I ) v 2 The value of ( A − I ) 2 can be obtained as, ( A − I ) 2 = [ 0 0 0 − 2 − 3 − 3 2 3 3 ] [ 0 0 0 − 2 − 3 − 3 2 3 3 ] = [ 0 0 0 0 0 0 0 0 0 ] Therefore, any vector v ≠ 0 is a solution of the equation ( A − I ) 3 v = 0 . Consider the vector v 2 = [ 1 0 0 ] Now associated vector v 1 can be obtained as, v 1 = [ 0 0 0 − 2 − 3 − 3 2 3 3 ] [ 1 0 0 ] = [ 0 − 2 2 ] The Eigen vector v 1 is not equal to the two Eigen vectors u 1 and u 2

5th Edition

ISBN: 9780321974235

Author: Calvis

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Chapter 5.5, Problem 16P

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Program Description:Purpose of problem is to obtain the general solution of the system x′=[100−2−2−3234]x .

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Chapter 5.5, Problem 15P

Chapter 5.5, Problem 17P

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Chapter 5 Solutions

EBK DIFFERENTIAL EQUATIONS

Ch. 5.1 - Let A=[2347] and B=[3451]. Find (a) 2A+3B; (b)...Ch. 5.1 - Prob. 2P Ch. 5.1 - Find AB and BA given A=[203415] and B=[137032].Ch. 5.1 - Prob. 4P Ch. 5.1 - Prob. 5P Ch. 5.1 - Prob. 6P Ch. 5.1 - Prob. 7P Ch. 5.1 - Prob. 8P Ch. 5.1 - Prob. 9P Ch. 5.1 - Prob. 10P

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Program Description: Purpose of problem is to obtain the general solution of the system x ′ = [ 1 0 0 − 2 − 2 − 3 2 3 4 ] x .

Solution Summary: The author explains that the problem is to obtain the general solution of the system of equation as A. The Eigen vector for the coefficient matrix A is in the form of left[cc

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Program Description:Purpose of problem is to obtain the general solution of the system x′=[100−2−2−3234]x .

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Chapter 5 Solutions